Statistical thermodynamics 1: the concepts - W.H. Freeman

More documents

Recommendations

Info

$Microscopy Math - W.H. Freeman$

PC8eC16 1/26/06 14:34 Page 575 16.4 THE STATISTICAL ENTROPY 575 Then, by eqn 16.31a, A D A 3V D 3N U = U(0) − NB E B− E =U(0) + C V F C 2βΛ 3 F 2β as in eqn 16.32b. Λ 3 16.4 The statistical entropy If it is true that the partition function contains all thermodynamic information, then it must be possible to use it to calculate the entropy as well as the internal energy. Because we know (from Section 3.2) that entropy is related to the dispersal of energy and that the partition function is a measure of the number of thermally accessible states, we can be confident that the two are indeed related. We shall develop the relation between the entropy and the partition function in two stages. In Further information 16.2, we justify one of the most celebrated equations in statistical thermodynamics, the Boltzmann formula for the entropy: S = k ln W [16.34] In this expression, W is the weight of the most probable configuration of the system. In the second stage, we express W in terms of the partition function. The statistical entropy behaves in exactly the same way as the thermodynamic entropy. Thus, as the temperature is lowered, the value of W, and hence of S, decreases because fewer configurations are compatible with the total energy. In the limit T → 0, W = 1, so ln W = 0, because only one configuration (every molecule in the lowest level) is compatible with E = 0. It follows that S → 0 as T → 0, which is compatible with the Third Law of thermodynamics, that the entropies of all perfect crystals approach the same value as T → 0 (Section 3.4). Now we relate the Boltzmann formula for the entropy to the partition function. To do so, we substitute the expression for ln W given in eqn 16.3 into eqn 16.34 and, as shown in the Justification below, obtain U − U(0) S = +Nk ln q (16.35) T Justification 16.4 The statistical entropy The first stage is to use eqn 16.3 (ln W = N ln N −∑ i n i ln n i ) and N =∑ i n i to write S = k∑ i (n i ln N − n i ln n i ) = −k∑n i ln i = −Nk∑ p i ln p i i where p i = n i /N, the fraction of molecules in state i. It follows from eqn 16.7 that ln p i =−βε i − ln q and therefore that S =−Nk(−β∑ i p i ε i −∑ p i ln q) = kβ{U − U(0)} + Nk ln q i We have used the fact that the sum over the p i is equal to 1 and that (from eqns 16.27 and 16.30) N ∑ p i ε i =∑Np i ε i =∑Np i ε i =∑n i ε i = E = U − U(0) i i i i We have already established that β = 1/kT, so eqn 16.35 immediately follows. n i N
PC8eC16 1/26/06 14:34 Page 576 576 16 STATISTICAL THERMODYNAMICS 1: THE CONCEPTS SNk / 4 2 0 0 5 10 kT/ Fig. 16.12 The temperature variation of the entropy of the system shown in Fig. 16.3 (expressed here as a multiple of Nk). The entropy approaches zero as T → 0, and increases without limit as T →∞. Exploration Plot the function dS/dT, the temperature coefficient of the entropy, against kT/ε. Is there a temperature at which this coefficient passes through a maximum If you find a maximum, explain its physical origins. Example 16.4 Calculating the entropy of a collection of oscillators Calculate the entropy of a collection of N independent harmonic oscillators, and evaluate it using vibrational data for I 2 vapour at 25°C (Example 16.3). Method To use eqn 16.35, we use the partition function for a molecule with evenly spaced vibrational energy levels, eqn 16.12. With the partition function available, the internal energy can be found by differentiation (as in eqn 16.31a), and the two expressions then combined to give S. Answer The molecular partition function as given in eqn 16.12 is 1 q = 1 − e −βε The internal energy is obtained by using eqn 16.31a: N A ∂q D Nεe −βε Nε U − U(0) =− = = q C ∂β F V 1 − e −βε e −βε − 1 The entropy is therefore 1 βε 5 S = Nk 2 − ln(1 − e βε ) 6 3 e βε − 1 7 This function is plotted in Fig. 16.12. For I 2 at 25°C, βε = 1.036 (Example 16.3), so S m = 8.38 J K −1 mol −1 . Self-test 16.6 Evaluate the molar entropy of N two-level systems and plot the resulting expression. What is the entropy when the two states are equally thermally accessible [S/Nk = βε/(1 + e βε ) + ln(1 + e −βε ); see Fig. 16.13; S = Nk ln 2] 1 1 SNk / ln 2 SNk / ln 2 0.5 0.5 Fig. 16.13 The temperature variation of the entropy of a two-level system (expressed as a multiple of Nk). As T →∞, the two states become equally populated and S approaches Nk ln 2. Exploration Draw graphs similar to those in Fig. 16.13 for a three-level system with levels 0, ε, and 2ε. 0 0 0.5 kT/ 1 0 0 5 10 kT/
Page 1 and 2: PC8eC16 1/26/06 14:34 Page 560 16 S
Page 3 and 4: PC8eC16 1/26/06 14:34 Page 562 562
Page 15: PC8eC16 1/26/06 14:34 Page 574 574
Page 29: PC8eC16 1/26/06 14:34 Page 588 588

Statistical thermodynamics 1: the concepts - W.H. Freeman

Create successful ePaper yourself

Delete template?

Save as template?